Worked examples — Separation techniques — filtration, distillation, chromatography, centrifugation, sublimation

Before anything, four words that will appear constantly, each anchored to a picture you already know:

Distillation examples below compare two liquids' boiling points, so we need one more piece of shorthand up front, defined here before it is ever used:

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The scenario matrix

Every separation problem is really asking: "Which physical property differs, and does any trap apply?" The grid below lists every case class this topic can present. Each later example is tagged with the cell it fills.

Cell	Case class	The trap / twist it tests	Example
A	Insoluble solid + liquid	Straightforward size difference	Ex 1
B	Soluble solid + liquid	Filtration fails — solute passes	Ex 2
C	Solid dissolved in liquid (one boils, one does not)	Only one component vaporises	Ex 3
D	Two liquids, small ΔBP	Simple fails, need fractional	Ex 4
E	Degenerate: ΔBP → 0	Azeotrope, no full separation	Ex 5
F	Solid that vaporises directly	Sublimation, skips liquid	Ex 6
G	Density difference, no settling	Centrifugation	Ex 7
H	Two immiscible liquids	Liquid–liquid extraction, two layers	Ex 8
I	Many components, polarity spread	Chromatography, $R_f$ numbers	Ex 9
J	Real-world multi-step	Chain techniques in order	Ex 10
K	Exam twist: decomposition	"Distill it" trap — must refuse	Ex 11

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Cell A — Insoluble solid + liquid

1. Compare grain size to pore size. Grain 100 μm vs pore 5 μm. Why this step? Filtration is purely a size gate. Only if the particle is bigger than the hole is it trapped. $100>5$, so the chalk is caught.
2. Pour the mixture through filter paper in a funnel; let gravity pull the water down. Why this step? Water molecules (≈ 0.0003 μm) sail through; chalk grains cannot fit.
3. Scrape the wet solid off the paper (residue = chalk); the clear liquid below is the filtrate = water. Why this step? Physical collection finishes the job.
4. Mass check: all 8.0 g is retained because none of it was dissolved. Residue mass = 8.0 g.

Verify: Chalk is insoluble, so nothing dissolves into the water; the size gate keeps 100% of it. Residue = 8.0 g, recovered fraction = $8.0/8.0=1.00$. Units: grams in, grams out — mass conserved. ✓

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Cell B — Soluble solid + liquid (filtration must FAIL)

1. Ask what a dissolved ion actually is. Dissolved ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ ions are ≈ 0.0003 μm across — thousands of times smaller than a 5 μm pore. Why this step? Filtration only distinguishes by size. A dissolved solute is molecularly mixed with the water; there is no lump to catch.
2. Predict the outcome. Every ion passes into the filtrate along with the water. Why this step? This is the deliberate failure — it teaches when not to reach for a filter.
3. Residue on paper = 0 g. Filtrate = salt water, unchanged.

Verify: Compare with Ex 1: chalk 100 μm → trapped; salt-ion 0.0003 μm → passes. Same apparatus, opposite result, and the difference is only size relative to the pore. To actually get the salt you must evaporate or distil (Cell C / Ex 3). Residue = 0 g. ✓

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Cell C — Solid dissolved in liquid: only one component boils

1. Identify the volatile component. Water boils at 100 °C; salt cannot boil at any temperature. Why this step? Distillation separates by boiling point. Here ΔBP does not apply — salt has no boiling point to subtract — so this is not a "gap" problem at all. It is the cleanest possible case: only one of the two things can ever become vapour, so whatever collects in the condenser is pure water.
2. Heat to 100 °C. Water becomes vapour and rises into the condenser. Why this step? At its BP, the water's escaping-vapour push equals atmospheric pressure — see Vapor Pressure — so it boils away.
3. Cold water jacket cools the vapour below 100 °C; it condenses to liquid = distillate. Why this step? Removing heat reverses the phase change, giving back pure liquid water.
4. Salt stays as solid crust in the flask (the residue).

Verify: Distillate should be salt-free. Because only water can vaporise, the distillate is pure water regardless of any ΔBP — the 25 °C rule is irrelevant here (it governs two boiling liquids, and salt never boils). Mass out: 200 mL water recovered (minus small losses), 12 g salt left behind. ✓

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Cell D — Two liquids, SMALL ΔBP

The two boiling points sit close together on the temperature axis of the figure below — look how small the double-headed yellow arrow on the right is compared with the huge one on the left.

1. Compute the gap. $\Delta \text{BP}=78-65=13\text{°C}$ (the short yellow arrow in the figure). Why this step? Using our difficulty dial: $13<25$, so a single vaporise–condense cycle gives a mixed vapour, not pure methanol — simple distillation is not good enough.
2. Use a fractionating column (packed tube) between flask and condenser. Why this step? Each bead in the column is a mini re-boil: vapour condenses, re-evaporates, and each cycle enriches the more volatile methanol. Many cycles stacked act like an effectively much larger ΔBP.
3. Collect the fraction coming over near 65 °C first — methanol-rich distillate. Why this step? Methanol, boiling lower, dominates the early vapour; the thermometer at the top reads ≈ 65 °C while methanol distils, then climbs toward 78 °C as ethanol follows.

Verify: Decision rule: $\Delta \text{BP}=13\text{°C}<25\text{°C}\Rightarrow$ fractional needed. This is the small-arrow side of the figure; the large-arrow side (a pair of liquids hundreds of degrees apart) would need only simple distillation. Consistent. ✓

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Cell E — Degenerate limit: ΔBP → 0 (azeotrope)

1. Recall the trend. As you approach 95.6% ethanol, the vapour and the liquid have the same composition. Why this step? Distillation works only when vapour is richer in the volatile component than the liquid. When vapour composition = liquid composition, the enrichment per cycle is zero — the difficulty dial ΔBP has effectively gone to 0.
2. Interpret as a degenerate case. This constant-boiling mixture is an azeotrope: it boils at 78.1 °C as if it were a single pure substance. Why this step? No number of extra column cycles helps, because each cycle multiplies enrichment by zero.
3. Escape route: add a third agent (e.g. benzene) or use molecular sieves — chemistry beyond simple distillation.

Verify: This is exactly Cell D pushed to its limit: ΔBP $13\to 0$ ⇒ separation power $\to$ none. The wall is at 95.6% ethanol by volume (the standard literature figure — do not quote it as "by mass", which gives a different number ≈ 97.2%). Azeotrope boiling point 78.1 °C is below pure ethanol's 78.4 °C — the signature of a minimum-boiling azeotrope. ✓ (see Colligative Properties)

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Cell F — Solid that vaporises directly (sublimation)

In the figure below, follow the pink arrows: they run straight from the warm dish at the bottom to the cold plate at the top with no liquid stage in between — that missing middle is the whole point of sublimation.

1. Gently heat the mixture in a dish, cold surface held above it. Why this step? Sublimation exploits a solid that skips the liquid phase — see States of Matter. Iodine turns straight to purple vapour (the pink arrows); sand stays solid in the dish.
2. Vapour hits the cold surface and re-deposits as pure crystals (the blue line at the top of the figure). Why this step? Cooling reverses sublimation (deposition), leaving sand behind in the dish.
3. Scrape purified iodine off the cold surface. Residue = sand.

Verify: Only iodine sublimes, so 5 g iodine collects on the cold plate, 20 g sand remains. Recovery fraction = $5/5=1.00$; total mass $5+20=25$ g conserved. Iodine never passes through a liquid state. ✓

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Cell G — Density difference, no natural settling (centrifugation)

1. Note the tiny density gap $1.10-1.03=0.07\text{g/mL}$ and small cell size ⇒ gravity settling is far too slow. Why this step? Gravity's pull is weak; small dense particles fall painfully slowly.
2. Spin the tube fast in a centrifuge. Spinning creates a large outward (centrifugal) effect — effectively "gravity" thousands of times stronger. Why this step? A stronger effective field makes the denser cells migrate outward (to the tube bottom) in minutes instead of hours.
3. Read the layers: denser red cells at the bottom, lighter plasma on top; decant the plasma.

Verify: Denser component (1.10 > 1.03) goes outward/down — consistent with "denser sinks." Density gap 0.07 g/mL is positive, so a stable layering exists. ✓

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Cell H — Two immiscible liquids (liquid–liquid extraction)

1. Why two liquids form layers at all. Water is polar, hexane is nonpolar; "like dissolves like" fails between them, so they refuse to mix and stack as two layers (hexane floating on water) — see Intermolecular Forces. Why this step? Immiscibility is what makes the two phases separable at all — after shaking, they re-split into clean layers you can tap off with a separating funnel.
2. Write the partition balance. At equal 100 mL volumes with a 4 : 1 preference, at equilibrium the hexane holds 4 parts for every 1 part left in water — so iodine splits $4:1$ into hexane : water. Why this step? Extraction separates by relative solubility (affinity), not size or boiling point. The 4 : 1 ratio is the driving number.
3. Split the 0.50 g. Total parts $=4+1=5$. Into hexane: $\frac{4}{5}\times 0.50=0.40$ g. Left in water: $\frac{1}{5}\times 0.50=0.10$ g. Why this step? Turning the ratio into masses answers the actual question.
4. Drain the lower water layer; keep the hexane layer holding 0.40 g iodine.

Verify: Masses add back: $0.40+0.10=0.50$ g — nothing lost. Fraction extracted $=0.40/0.50=0.80=80\%$, which is "more than half," consistent with iodine preferring the hexane. ✓

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Cell I — Many components, spread of polarity (chromatography)

Read the figure bottom-to-top: the yellow line is how far the solvent climbed, and the pink and blue dots mark where each dye stopped — $R_f$ is just each dot's height divided by the yellow line's height.

1. Write the retention factor. $R_f=\frac{\text{distance moved by dye}}{\text{distance moved by solvent front}}$. Why this step? $R_f$ is a ratio, so it cancels out how long you ran the plate — it is a fingerprint of the molecule, not the experiment.
2. Plug in dye X: $R_f(X)=\frac{6.0}{8.0}=0.75$ (the higher, pink dot). Why this step? Big number ⇒ travelled far ⇒ spent most time in the moving solvent ⇒ weakly stuck ⇒ less polar.
3. Plug in dye Y: $R_f(Y)=\frac{2.0}{8.0}=0.25$ (the lower, blue dot). Why this step? Small number ⇒ dragged along by the polar paper ⇒ more polar. Polar-likes-polar — see Intermolecular Forces.

Verify: Both $R_f$ values lie in $[0,1]$ as they must (a dye can never outrun the solvent front). $0.25<0.75$, so Y is the more polar, slower dye. ✓

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Cell J — Real-world multi-step chain

1. Filter first to remove insoluble silt (Cell A logic). Why this step? Silt is a solid lump — cheap filtration handles it. Distilling muddy water would just bake mud onto the flask; remove bulk solids first.
2. Distil the filtrate to separate water from dissolved salt (Cell C logic). Why this step? Salt is dissolved, so filtration can't touch it (Cell B). Only a boiling-point method leaves salt behind.
3. Collect distillate = pure water; residue trail = silt on paper + salt crust in flask.

Verify: Order test — reversing (distil first) wastes energy boiling mud and fouls glassware, so filter-then-distil is correct. Two property differences (size, then boiling) needed two techniques. ✓

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Cell K — Exam twist: refuse to distil a decomposer

1. Check whether the target can vaporise before it decomposes. Sucrose caramelises/decomposes at ≈ 186 °C — it never reaches a clean boiling point. Why this step? Distillation requires the component to vaporise intact. If it breaks down first, there is no vapour to condense — the method is invalid. (This is also why ΔBP is meaningless here: sugar has no boiling point, just like salt in Cell C.)
2. Reject distillation of the sugar. Instead evaporate the water gently, leaving sugar crystals behind (crystallisation). Why this step? We want the water to leave, not the sugar; evaporating water below 100 °C keeps the sugar safely solid rather than charred.
3. State the outcome. Gentle evaporation drives off the water as vapour; the sugar stays behind as recovered crystals. The correct answer to the exam question is therefore: the request is impossible as phrased — vaporise the water, not the sugar.

Verify: Decomposition temp 186 °C is reached before any hypothetical sugar-vaporisation temperature, so distilling the sugar is impossible; the water (BP 100 °C) is the component that should leave. Logic consistent with the parent note's decomposition warning. ✓

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Recall Self-test — cover the answers

Filter salt water: how much salt in residue? ::: 0 g — dissolved ions pass through the pores. ΔBP = 13 °C: simple or fractional distillation? ::: Fractional (below the ~25 °C practical line for simple). Why is ΔBP irrelevant for separating salt from water? ::: Salt has no boiling point (it decomposes), so there is no gap to compare — only water boils. $R_f$ of a dye that moves 3 cm when the front moves 12 cm? ::: $3/12=0.25$. Two immiscible liquids, iodine prefers hexane 4:1 — fraction extracted into hexane? ::: $4/5=0.80=80\%$. Denser blood cells after spinning sit where? ::: At the bottom (outward end) of the tube. Iodine + sand — which technique and does iodine melt? ::: Sublimation; iodine goes solid→gas, never melts. Why can't you distil sugar out of sugar water? ::: Sugar decomposes (~186 °C) before it could vaporise.